Question

Using the definition of derivative, how do you prove that (cos x)' = -sin x?

Answer 1

$\textsf{\underline {\Large {Derivative of cos x}}}$ :

$\textsf{\underline {\large {Proof}}}$ :

By using First Principle,

$\mathsf{f'(\:x\:) \:=\: {\displaystyle \lim_{h\to 0} {\dfrac{f(\:x\:+\:h\:) \:-\:f(\:x\:)}{h}}}}$

$\mathsf{f'(\:x\:) \:=\: {\displaystyle \lim_{h\to 0}{\dfrac{cos(\:x\:+\:h\:) \:-\:cos\:x}{h}}}}$

$\mathsf{f'(\:x\:) \:=\: {\displaystyle \lim_{h\to 0}{\dfrac{1}{h}[-2\:sin{\dfrac{(\:x\:+\:h\:+\:x)}{2}sin{\dfrac{(\:x\:+\:h\:-\:x)}{2}}}}}}]$

$\mathsf{f'(\:x\:) \:=\: {\displaystyle \lim_{h\to 0} {\dfrac{1}{h} [- \:2\:sin(\:x\:+\:{\dfrac{h}{2})\:sin\:({\dfrac{h} {2}}}}}})]$

$\mathsf{f'(\:x\:) \:=\: {\displaystyle \lim_{h\to 0} [- \:2\:sin(\:x\:+\:{\dfrac{h}{2})\:*\:{\displaystyle \lim_{\dfrac{h}{2}\to 0}sin\:{\dfrac{\dfrac{h}{2}}{\dfrac{h}{2}}}}}}}$

$\mathsf{f'(\:x\:) \:=\:-\:sin\:(\:x\:+\:0\:)\:*\:1}$

$\boxed{\mathsf{f'(\:x\:) \:=\:-\:sin\:x}}$

$\textsf{\underline {\large{Formulae\:Used}}}$ :

⚫$\boxed{\mathsf{cos(\:A\:+\:B\:)\:=\:-2\:sin{\dfrac{(\:A\:+\:B)}{2} sin{\dfrac{(\:A\:-\:B)}{2}}}}}$